| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with elastic strings/springs |
| Difficulty | Standard +0.3 This is a standard M2 elastic strings question requiring routine application of Hooke's law for tensions and elastic PE, followed by energy conservation and Newton's second law. All steps follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02d Mechanical energy: KE and PE concepts6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \([T_A = 4 \times 0.75/0.25\ \text{and}\ T_B = 8 \times 0.75/0.25]\) | M1 | For using \(T = \lambda x/L\) |
| Tensions are \(12\ \text{N}\) and \(24\ \text{N}\) | A1 [2] | |
| (ii) Total \(EE = 4 \times 0.75^2/(2 \times 0.25) + 8 \times 0.75^2/(2 \times 0.25)\) | M1 | For using \(T = \lambda x^2/2L\) |
| Total \(EE = 13.5\ \text{J}\) | A1 [2] | AG |
| (iii) \([T_B - T_A = ma]\) | M1 | For using Newton's second law |
| Acceleration is \(7.5\ \text{ms}^{-2}\) | A1\(\sqrt{}\) [2] | Ft \(0.625(T_B - T_A)\) |
| (iv) \(4(0.75+x)^2/(2\times0.25) + 8(0.75-x)^2/(2\times0.25) = 13.5\) | M1, A1 | For attempting to set up an equation using EE |
| \([-12x(1-2x) = 0 \Rightarrow x = 0,\ \tfrac{1}{2}]\) | M1 | For attempting to solve the correct quadratic equation |
| Value of \(x\) is \(0.5\) | A1 [4] |
## Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $[T_A = 4 \times 0.75/0.25\ \text{and}\ T_B = 8 \times 0.75/0.25]$ | M1 | For using $T = \lambda x/L$ |
| Tensions are $12\ \text{N}$ and $24\ \text{N}$ | A1 [2] | |
| **(ii)** Total $EE = 4 \times 0.75^2/(2 \times 0.25) + 8 \times 0.75^2/(2 \times 0.25)$ | M1 | For using $T = \lambda x^2/2L$ |
| Total $EE = 13.5\ \text{J}$ | A1 [2] | AG |
| **(iii)** $[T_B - T_A = ma]$ | M1 | For using Newton's second law |
| Acceleration is $7.5\ \text{ms}^{-2}$ | A1$\sqrt{}$ [2] | Ft $0.625(T_B - T_A)$ |
| **(iv)** $4(0.75+x)^2/(2\times0.25) + 8(0.75-x)^2/(2\times0.25) = 13.5$ | M1, A1 | For attempting to set up an equation using EE |
| $[-12x(1-2x) = 0 \Rightarrow x = 0,\ \tfrac{1}{2}]$ | M1 | For attempting to solve the correct quadratic equation |
| Value of $x$ is $0.5$ | A1 [4] | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{fb79f949-567c-4dbb-8533-7b7278cad21c-3_200_639_1754_753}
A particle $P$ of mass 1.6 kg is attached to one end of each of two light elastic strings. The other ends of the strings are attached to fixed points $A$ and $B$ which are 2 m apart on a smooth horizontal table. The string attached to $A$ has natural length 0.25 m and modulus of elasticity 4 N , and the string attached to $B$ has natural length 0.25 m and modulus of elasticity 8 N . The particle is held at the mid-point $M$ of $A B$ (see diagram).\\
(i) Find the tensions in the strings.\\
(ii) Show that the total elastic potential energy in the two strings is 13.5 J .\\
$P$ is released from rest and in the subsequent motion both strings remain taut. The displacement of $P$ from $M$ is denoted by $x \mathrm {~m}$. Find\\
(iii) the initial acceleration of $P$,\\
(iv) the non-zero value of $x$ at which the speed of $P$ is zero.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fb79f949-567c-4dbb-8533-7b7278cad21c-4_529_542_269_804}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A uniform solid body has a cross-section as shown in Fig. 1.\\
(i) Show that the centre of mass of the body is 2.5 cm from the plane face containing $O B$ and 3.5 cm from the plane face containing $O A$.\\
(ii) The solid is placed on a rough plane which is initially horizontal. The coefficient of friction between the solid and the plane is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fb79f949-567c-4dbb-8533-7b7278cad21c-4_332_469_1320_918}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The solid is placed with $O A$ in contact with the plane, and then the plane is tilted so that $O A$ lies along a line of greatest slope with $A$ higher than $O$ (see Fig. 2). When the angle of inclination is sufficiently great the solid starts to topple (without sliding). Show that $\mu > \frac { 5 } { 7 }$.\\[0pt]
[5]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{fb79f949-567c-4dbb-8533-7b7278cad21c-4_291_465_1987_918}
Instead, the solid is placed with $O B$ in contact with the plane, and then the plane is tilted so that $O B$ lies along a line of greatest slope with $B$ higher than $O$ (see Fig. 3). When the angle of inclination is sufficiently great the solid starts to slide (without toppling). Find another inequality for $\mu$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2009 Q6 [10]}}